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Bending Stresses in Beams

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In this stack, our goal is to developa means for determining thestresses in a beam.

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We will proceed by first determining thestrains due to bending…

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…and then use Hooke's law to determinethe stresses.

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Beam

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  To begin our detailed look

at the deformations of a bent beam, consider a beam with asymmetric cross section.

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Centroid of Section

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  We can calculate thelocation of the centroid bysetting the first moment ofarea to zero. The centroidwill always lie on the axis ofsymmetry.

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Neutral Axis!

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  We define the neutral axis 

of the beam as a line whichexperiences no strain as the beam is bent. As we willdemonstrate at the end of thisstack, the neutral axis passesthrough the centroid of thesection at any point along the beam.

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Plane Section

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  Let's now look at the material which lies in aplane passed through the beam. This particularplane is normal to the neutral axis.

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Apply Load

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  We now load the beamand allow it to deflect.

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Deformed Beam

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  After deformations, we observe that the plane section we were viewing remainsplane, and further, it remains normal to the neutral axis.

This observation is the fundamental assumption in the derivation of the beam

 bending equations. Note that we will not observe this same behavior for very largedeformations.

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Two Adjacent Sections

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  Let's take another look at the deformation ofthe loaded beam, only this time we will look at thematerial which lies between two adjacent planes.

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After Loading

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  As we have observed before, the sectionsremain plane and normal to the neutral axisafter deformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

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  As we have assumed, the sections remainplane and normal to the neutral axis afterdeformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

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  As we have assumed, the sections remainplane and normal to the neutral axis afterdeformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

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  As we have assumed, the sections remainplane and normal to the neutral axis afterdeformation.

  For clarity, we will continue by looking at aside view of the deformed beam.

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  What we are really interested in is the displacedshape of the element lying between the two planes. Wefocus in on this particular element.

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  This is the displaced shape of theelement after loading. Note that thetwo planes defining the element arenormal to the neutral axis.

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  Here the element is shown before and after deflection.

How can we relate theposition of a point in thematerial before deformation to

the position of the same pointafter deformation?

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  In order to rigorously

define how a point in thematerial moves when the beam is loaded, we must firstdefine a coordinate system.  Here we define the x axisto run along the neutral axis.We also assume that thelength of the beam element issome value dx.

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  The distance of the pointabove or below the x axis(neutral axis) we will defineas "y". y is positive whenthe point lies above theneutral axis.

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  The displacement of theneutral axis from the

undeformed to the deformedconfiguration is described bythe function !  (x).

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  If ! " x) is thedisplacement of the beam at any point, x,then the firstderivative of thedisplacement,! ' (x) ,is the slope of the beam at the point x.

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  Let's compare ! ' (x) (the slope) of one side

of the deformed elementto the other. To do thiswe extend theundeformed orientationof the two planes downonto the deformedelement.

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  At this point we need toassume that the left face ofthe element is located a

distance x along the beam.From this assumption wecalculate that the right faceof the element is located adistance x + dx along the beam.

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  The angle between theplane defining the left faceof the deformed element

and its undeformedorientation is simply theslope of the beam at thatpoint, !  '(x).

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  Similarly, the angle between the plane definingthe right face of thedeformed element and itsundeformed orientation isthe slope of the beam at thatpoint, ! ' (x + dx)  .

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  We can expand thisexpression as shown below.We will neglect the higher

order terms since we aredealing with smalldisplacements, and thereforesmall changes in slopes.

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  The previous few steps mayhave been a bit confusing, solet's go through them againwith a bigger picture.

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  We begin by projectingthe deformed element ontothe undeformed element.

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  We were able tocalculate the angle between theundeformed anddeformed planes asshown above.

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  Recall that the length of an arc

may be calculated as the angledefining the arc times the radiusof the arc.  Using this knowledge we areable to calculate the horizontaldisplacement of the upper-leftcorner of the element as:  y! ' (x)  .

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  We can calculate thedisplacement of theupper-right corner of theelement in a similar fashion.

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  The change in length of thetop chord of the element can

now be calculated as thedifference between thedisplacement of the upper-leftand upper-right corners.

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  Eliminating like terms, wefind that the change in lengthof the top chord is given by:

 – y! '' (x)  dx  

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  Recall that strain is

calculated as change in lengthdivided by original length.

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The strain of the material at the topof the element is then calculated asthe change in length of the topchord divided by the originallength of the top chord.

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  Note that we have actually kept the location where we calculate the strain in terms

of the y coordinate. Our equation tells us that when y is zero there is no strain. Wecan confirm this by noting that the element does not change length at the neutral axis.

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  Now that we have anexpression for the strain at anypoint in the beam, how do wecalculate the stress?

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  Hooke's Law tells us thatstress is linearly related to strain by the material constant, E.

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  Combining the two equationswe can relate the stress at a pointin the beam to the displacedshape of the beam and Young'sModulus.

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  Plotting the stress on the

element, we see that it varieslinearly with y , the distancefrom the neutral axis. Also, ifthe function! '' (x)  (thecurvature of the beam) is notconstant, then the stressvaries along the length of the beam as well.

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  Since it is not common thatwe begin with an expressionfor the displaced shape of the beam, let's try to manipulateour equation for stress due to bending until it is more useful.  Here we introduce thenotion that internal moment,M, is actually the sum of manysmaller moments caused by

stress acting away from theneutral axis.

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  Next, we substitute ourexpression for stress.

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  Rearranging terms...

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  Recall the expression for

moment of inertia.

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  Performing thesubstitution we find thatinternal moment is related to

the curvature of the beam.

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  How can we use this resultto simplify our expression forstress in the beam?

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  We use the secondexpression to eliminate ! '' (x) from the stress equation and wearrive at the famousrelationship:

  " = –M y/ I

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  Here is a summary of the importantrelationships we have derived usingthe displacement assumption that"plane sections remain plane, andnormal to the neutral axis."

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So far we have worked on the assumptionthat the neutral axis and the centroidal axiscoincide. We will now demonstrate thevalidity of this assumption. To this end,

consider the net horizontal force, P.

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The net force can be calculated by integrating thestresses over the cross section. Since no horizontalloads have been applied to the beam, the net force

must be zero.

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Using our previous results, we canexpress the stresses in terms of thedisplacement and material

properties as shown.

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We can pull E and ! '' out of theintegral, since they do not vary

over the cross-section.

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Location of Neutral Axis

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Since E and ! '' are notzero in general, the

 boxed result must hold.This is simply thedefinition of the centroid,and so the proper placefrom which y ismeasured is the centroid.Thus, the neutral axis andthe centroidal axis arecoincident.

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Summary

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  The relation between bending moment and the resulting

stresses is extremely important, and you are likely to encounterit again and again. You should store " = My/I somewhere inyour brain near F = ma.  Remember, the linear distribution of stress predicted bythis equation is based on the assumed "plane sections remainplane and normal to the neutral axis" assumption, which is anapproximation (but a darn good one as long as the beam's

length is more than about 3 to 4 times its depth).

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End

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